Question: Factor the quadratic expression completely. $15x^2-4x-4=$
Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${15}x^2{-4}x{-4}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={-4} \\\\ &{ab}={A}{C}= ({15})({-4})=-60 \end{cases}$ We find that ${a}={6}$ and ${b}={-10}$ satisfy these conditions, since ${6}+({-10)}={-4}$ and $({6})({-10})=-60$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 15x^2-4x-4&=15x^2+{6}x{-10}x-4 \\\\ &=3x(5x+2)-2(5x+2) \\\\ &=(5x+2)(3x-2) \end{aligned}$ In conclusion, $15x^2-4x-4=(5x+2)(3x-2)$